During complete combustion of one mole of butane ,2658Kj of heat is released. The thermochemical reaction for above change is

Options

A. `2C_(4) H_(10) (g) + 13 O_(2) (g) rarr 8 CO_(2)(g) + 10 H_(2) O (l), Delta_(c) H = -2658.0 kJ mol^(-1)`
B. `C_(4)H_(10) (g) + (13)/(2) O_(2) (g) rarr 4CO_(2) (g) + 5H_(2)O (l) , Delta_(c) H = - 1329.0 kJ mol^(-1)`
C. `C_(4)H_(10) (g) + (13)/(2) O_(2)(g) rarr 4CO_(2) (g) + 5H_(2) O (l), Delta_(c) H = - 2658.0 kJ mol^(-1)`
D. `C_(4) H_(10) (g) + (13)/(2) O_(2) (g) rarr 4CO_(2) (g) + 5H_(2) O (l) , Delta_(c) H = + 2658.0 kJ mol^(-1)`

Answer

C

Explanation

Give that, the complete combustion of one mole of butane is represented by thermochemical reaction as
`C_(4) H_(10) (g) + (13)/(2) O_(2) (g) rarr 4CO_(2) (g) + 5H_(2) O (l)`
We have to take the combustion of one mole of `C_(4)H_(10)` and `Delta_(c)H` should be negative and have a value of `2658 kJ mol^(-1)`